Divisors of One More than Power of 10/Number of Zero Digits Even

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Theorem

Let $N$ be a natural number of the form:

$N = 1 \underbrace {000 \ldots 0}_{\text {$2 k$ $0$'s} } 1$

that is, where the number of zero digits between the two $1$ digits is even.

Then $N$ can be expressed as:

$N = 11 \times \underbrace {9090 \ldots 90}_{\text {$k - 1$ $90$'s} } 91$


Proof

By definition, $N$ can be expressed as:

$N = 10^{2 k + 1} + 1$

Then we have:

\(\ds N\) \(=\) \(\ds 10^{2 k + 1} + 1\)
\(\ds \) \(=\) \(\ds \paren {10 + 1} \sum_{j \mathop = 0}^{2 k} \paren {-1}^j 10^{2 k - j}\) Sum of Two Odd Powers
\(\ds \leadsto \ \ \) \(\ds \dfrac N {11}\) \(=\) \(\ds \sum_{j \mathop = 0}^{2 k} \paren {-1}^j 10^{2 k - j}\)
\(\ds \) \(=\) \(\ds 10^{2 k} - 10^{2 k - 1} + 10^{2 k - 2} - 10^{2 k - 3} + \cdots + 10^2 - 10^1 + 10^0\)
\(\ds \) \(=\) \(\ds 10^{2 k - 1} \paren {10 - 1} + 10^{2 k - 3} \paren {10 - 1} + \cdots + 10 \paren {10 - 1} + 1\)
\(\ds \) \(=\) \(\ds 90 \paren {10^{2 k - 2} + 10^{2 k - 4} + \cdots + 10^0} + 1\)
\(\ds \) \(=\) \(\ds 90 \sum_{j \mathop = 0}^k \paren {10^{2 j} } + 1\)
\(\ds \) \(=\) \(\ds \underbrace {9090 \ldots 90}_{\text {$k$ $90$'s} } + 1\)
\(\ds \) \(=\) \(\ds \underbrace {9090 \ldots 90}_{\text {$k - 1$ $90$'s} } 91\)

$\blacksquare$


Examples

\(\ds 11\) \(=\) \(\ds 11\)
\(\ds 1001\) \(=\) \(\ds 11 \times 91\)
\(\ds \) \(=\) \(\ds 7 \times 11 \times 13\)
\(\ds 100 \, 001\) \(=\) \(\ds 11 \times 9091\)
\(\ds 10 \, 000 \, 001\) \(=\) \(\ds 11 \times 909 \, 091\)
\(\ds 1 \, 000 \, 000 \, 001\) \(=\) \(\ds 11 \times 90 \, 909 \, 091\)
\(\ds \) \(=\) \(\ds 7 \times 11 \times 13 \times 19 \times 52 \, 579\)
\(\ds 100 \, 000 \, 000 \, 001\) \(=\) \(\ds 11 \times 9 \, 090 \, 909 \, 091\)
\(\ds \) \(=\) \(\ds 11^2 \times 23 \times 4093 \times 8779\)


Also see


Sources

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